Problem: Is the function given below continuous/differentiable at $x=2$ ? $f(x)=\begin{cases} x^2+1&,&x\leq 2 \\\\ x-x^2&,&x>2 \end{cases}$ Choose 1 answer: Choose 1 answer: (Choice A) A Continuous but not differentiable (Choice B) B Differentiable but not continuous (Choice C) C Both continuous and differentiable (Choice D) D Neither continuous nor differentiable
Explanation: Checking for continuity at $x=2$ For the function to be continuous at $x=2$, we need the two-sided limit $\lim_{x\to 2}f(x)$ to exist and be equal to $f(2)$. This is the same as requiring that the two one-sided limits $\lim_{x\to 2^-}f(x)$ and $\lim_{x\to 2^+}f(x)$ exist and are equal to $f(2)$. According to $f$ 's definition, $f(2)=(2)^2+1=5$. $\lim_{x\to 2^-}f(x)$ : $x^2+1$ evaluated at $x=2$ is equal to $5$. Since $x^2+1$ is continuous, we can be certain that $\lim_{x\to 2^-}f(x)=5$. $\lim_{x\to 2^+}f(x)$ : $x-x^2$ evaluated at $x=2$ is equal to $-2$. Since $x-x^2$ is continuous, we can be certain that $\lim_{x\to 2^+}f(x)=-2$. The two limits exits, but they are not equal. Therefore, the function is not continuous at $x=2$. Graphically, the function skips a step at this point. [I would like to see that, please!] Checking for differentiability at $x=2$ Since the function isn't continuous at $x=2$, it cannot be differentiable at that point. In conclusion, the function is neither continuous nor differentiable at $x=2$.